On average, a sharpshooter hits the target once every 3 shots. What

Question

On average, a sharpshooter hits the target once every 3 shots. What is the probability that he will hit the target in 4 shots?

(A) 1
(B) 1/81
(C) 1/3
(D) 65/81
(E) 71/81

eswarchethu135 · Accepted Answer

P(hitting the target) =  \frac{1}{3}

P(not hitting the target) =  \frac{2}{3}

P(hitting the target in 4 shots) = 1 - P(not hitting the target in 4 shots)

=  1 - (\frac{2}{3}*\frac{2}{3}*\frac{2}{3}*\frac{2}{3})

=  1 - \frac{16}{81}

=  \frac{65}{81}

OPTION:  D

Archit3110 · Answer

hiting target 1/3
missing 1-1/3 ; 2/3
so in 4 shots
1-(2/3)^3 
1-16/81
65/81
IMO D

neha283 · Answer

Hi, i have a doubt, according to the question, "On average, a sharpshooter hits the target once every 3 shots". So by 3rd shot, he must have hit once and by 4th shot too he must have hit atleast once. Why is the answer not 1 then?

Lobna12 · Answer

I got B.

(1/3)^4=1/81. Can anyone help why this is not correct? Appreciate it.

neha283 · Answer

The approach that you have taken means the shooter will hit in every shot, which is not what the question is asking.

Hope it helps.

HamedATX · Answer

you can rephrase the question this way:
What is the probability that a sharpshooter AT LEAST hit one target in 4 shots.

Hope it helps.

wadhwakaran · Answer

i think it should be 1. experts please shed light on this

Regor60 · Answer

The first response to this question provides the clearest answer.

The AVERAGE probability is 1/3.

That means that over a series of many shots a hit was made in 1 shot, 2 shots, 20 shots, etc, averaging a success rate of 1 in 3.

Because some hits were made only after many shots, for example, it should be clear that there is no guarantee that a shot will be made in ANY series of shots.

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ThatDudeKnows · Answer

@wadwakaran
You are ignoring the first two words of the question: "on average." A fair coin lands on heads half the time and on tails half the time, so  on average , it lands on heads once every 2 tosses. Does that mean that if you toss the coin twice, you will definitely get one heads and one tails? Nope. You could get heads 100 times in a row. Or tails 100 times in a row. Similarly, it is entirely possible for the sharpshooter to hit the target four times in a row or to miss it four times in a row.

I find the distribution of answer choices selected by forum members to be really interesting on this one.
18% A
12% B
12% C
56% D
2% E

Only 56% got it right, so let's figure out WHY!

Without doing any math and just by reasoning through this, we should be able to get it down to D and E, so the fact that > 40% chose A, B, or C is a problem!

A) As above, just as it is possible to get four heads in a row on coin tosses, it's possible for the sharpshooter to miss the target all four times. Incorrect.

B and C) If his odds are 1/3 on a single shot, should taking more shots improve his odds, keep them the same, or cause them to diminish? Let's use some logic. If we gave him 10000000000 shots, he'd eventually hit it, right? If we gave him 10000 shots, he'd still be pretty sure of hitting it. If we gave him 100 shots, his odds would still be pretty good. Even with 10 shots, the odds would certainly be better than 1/3. So, having more than one shot means his odds are higher than 1/3. If you picked B or C, please slow down and make sure that you aren't throwing calculations at a problem before you really think about what you're calculating. If you have no idea how to do a problem, see if you can reason through it like this to eliminate answer choices. We haven't done any math at all, we are down to a 50/50 with D and E...and we've eliminated the three wrong answers that people picked 40% of the time.

Alright, so now let's work through the math. The calculation in the first response on the thread is correct, but it's pretty clear from the fact that so many folks are getting this wrong that they don't understand WHY.

On probability questions, the first thing we need to do is define "how to win." We win with any of the following:
Hit and hit and hit and hit...or
Hit and hit and hit and miss...or
Hit and hit and miss and hit...or
Hit and miss and hit and hit...or
Miss and hit and hit and hit...or
Hit and hit and miss and miss...or
Hit and miss and hit and miss...or
Hit and miss and miss and hit...or
Miss and hit and hit and miss...or
Miss and hit and miss and hit...or
Miss and miss and hit and hit...or
Hit and miss and miss and miss...or
Miss and hit and miss and miss...or
Miss and miss and hit and miss...or
Miss and miss and miss and hit

Oh, dear, that's a LOT of stuff to list out...but I have great news!!! The odds of winning and the odds of losing add up to be 100%. W+L=1. That means that instead of doing all the work to find the odds of winning, we can just find the odds of losing and then use W=1-L.

Remember when I said above that the first step on probability questions is defining how we win? Well, I kinda take it back. The first step on probability questions is defining EITHER how we win OR how we lose. In this case, how we lose is much easier. It's just:

Miss and miss and miss and miss.

In probability, "and" means multiply and "or" means "add."
We know from the question that the probability of a hit on each individual shot is 1/3. So, what's the probability of a miss on each individual shot? 2/3.

Okay, so let's plug in "2/3" for each "miss" and "*" for each "and."

Miss and miss and miss and miss
2/3 * 2/3 * 2/3 * 2/3 = 16/81

So that's the probability that the sharpshooter MISSES ALL FOUR TIMES. But that's not a win; we defined winning as getting at least one hit, so missing all four times is a LOSS.

W+L=1
W + (16/81) = 1
W = 1 - (16/81)
W = 65/81

Answer choice D.

ThatDudeKnowsProbability

Zareensoobuh · Answer

Can we do this math by using combination method?

Posted from my mobile device

Bunuel · Answer

I'd recommend to stick with  this method .

On average, a sharpshooter hits the target once every 3 shots. What

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On average, a sharpshooter hits the target once every 3 shots. What

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